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Let $f(x)$ be continuous on $R$. Suppose that $f$ is periodic with the minimal postive period $mu>0$, $mu$ is irrational). Show that $lim_{ntoinfty} f(n)$ does not exist.

I do know that ${n+mmu; n,min Z}$ is dense, but for $x=lim_{ktoinfty}(n_k+m_kmu)$, I could not ensure that $n$ is positive, and so could not get a contradiction with the non-existence of $lim_{ntoinfty} f(n)$.

A related fact is that ${0+mmu mod 1}_{min mathbb{N}}$ is dense in $[0,1)$. This means (check this) there is a subsequence $m_k uparrow infty$ such that

$$
{m_k mu - [m_kmu] downarrow 0}
$$
where $[x]$ denotes the integer part.
Call $n_k:= [m_kmu]$ and note $n_krightarrow infty$.

And $f(n_k) = f(m_kmu-epsilon_k) = f(-epsilon_k) rightarrow f(0)$ by continuity, where $epsilon_k$ is a quantity going to 0.

In fact, we found subsequence conveging to $f(0)$ but you can find a subsequence converging to $f(x_0)$ for any $x_0$. This is because every orbit of $+mu$ on $[0,1)$ is dense, i.e., for any $x_0$, the set ${x_0+mmu mod 1}_{minmathbb{N}}$ is dense in $[0,1)$. You can find a subsequence $m_juparrow infty$ such that
$$x_0+m_jmu - [x_0+m_jmu] downarrow 0$$ and repeating as above, go on to show that $f(n_j) rightarrow f(x_0)$.